Question

question 15 (3 points)

a flare is launched from a launch pad straight up into the air. the height of the flare, h, in metres is modelled by \\(h(t) = -5t^2 + 100t + 5\\) where \\(h(t)\\) is the height of the flare above the ground, in metres, and \\(t\\) is the time the flare is in the air, in seconds.

a. what height was the flare launched from?

b. what is the maximum height of the flare above the ground and how long does it take to reach the maximum height?

Explanation:

Evaluate the initial height

To find the launch height, evaluate the height function at the initial time \(t = 0\).

$$ h(0) = -5(0)^2 + 100(0) + 5 = 5 $$

Find the time to reach maximum height

The maximum height of a quadratic function \(h(t) = at^2 + bt + c\) occurs at its vertex.
Using the Vertex of a Quadratic Function formula for the time coordinate \(t\):

$$ t = -\frac{b}{2a} = -\frac{100}{2(-5)} = 10\text{ seconds} $$

Calculate the maximum height

Substitute the vertex time \(t = 10\) back into the height function to find the maximum height.
Alternatively, we can use Completing the Square to rewrite the function:

$$ h(t) = -5(t^2 - 20t) + 5 = -5(t - 10)^2 + 505 $$

Thus, the maximum height is \(505\text{ metres}\).

Answer:

Question a

The flare was launched from a height of 5 metres.

Question b

The maximum height of the flare is 505 metres, and it takes 10 seconds to reach this height.